Question 6.5: Newton's interpolating polynomial. The set of the following ......
Newton’s interpolating polynomial.
The set of the following five data points is given:
\begin{array}{cr}x & 1 & 2 & 4 & 5 & 7 \\y & 52 & 5 & -5 & -40 & 10\end{array}(a) Determine the fourth-order polynomial in Newton’s form that passes through the points. Calculate the coefficients by using a divided difference table.
(b) Use the polynomial obtained in part (a) to determine the interpolated value for x=3.
(c) Write a MATLAB user-defined function that interpolates using Newton’s polynomial. The input to the function should be the coordinates of the given data points and the x coordinate of the point at which y is to be interpolated. The output from the function is the y value of the interpolated point.
(a) Newton’s polynomial for the given points has the form:
f(x)=y=a_{1}+a_{2}(x-1)+a_{3}(x-1)(x-2)+a_{4}(x-1)(x-2)(x-4)+a_{5}(x-1)(x-2)(x-4)(x-5)With the coefficients determined, the polynomial is:
f(x)=y=52-47(x-1)+14(x-1)(x-2)-6(x-1)(x-2)(x-4)+2(x-1)(x-2)(x-4)(x-5)(b) The interpolated value for x=3 is obtained by substituting for x in the polynomial:
f(3)=y=52-47(3-1)+14(3-1)(3-2)-6(3-1)(3-2)(3-4)+2(3-1)(3-2)(3-4)(3-5)=6
Script File
(c) The MATLAB user-defined function for Newton's interpolation is named Yint=NewtonsINT (x, y, X int ) . x and y are vectors with the coordinates of the given data points, and Xint is the coordinate of the point at which y is to be interpolated.
- The program starts by calculating the first divided differences, which are then used for calculating the higher divided differences. The values are assigned to a table named divDIF.
- The coefficients of the polynomial (first row of the table) are then assigned to a vector named a.
- The known polynomial is used for interpolation.
Program 6-5: User-defined function. Interpolation using Newton's polynomial.
function Yint = NewtonsINT(x,y,Xint)
% NewtonsINT fits a Newtons polynomial to a set of given points and
% uses the polynomial to determines the interpolated value of a point.
% Input variables:
% x A vector with the x coordinates of the given points.
% y A vector with the y coordinates of the given points.
% Xint The x coordinate of the point to be interpolated.
% Output variable:
% Yint The interpolated value of Xint.
n = length(x) ;
a (l)= y (l);
for i = l: n - 1
divDIF(i,l)=(y(i+l)-y(i))/(x(i+ 1)-x(i));
end
for j = 2 :n - 1 for i = l:n - j
divDIF(i,j)=(divDIF(i+l,j-l) -divDIF(i,j-1))/(x(j+i)-x(i));
end
end
for j = 2 :n
a(j) = divDIF(l,j - 1); end
Yint =a (1);
xn = 1;
for k = 2:n
xn = xn*(Xint - x (k - 1));
Yint = Yint + a (k) *xn;
end
The length of the vector \mathrm{x} gives the number of coefficients (and terms) of the polynomial
The first coefficient a_{1}.
Calculate the finite divided differences. They are assigned to the first of divDIF.
Calculate the second and higher divided differences (up to an order of (n-1) ). The values are assigned in columns to divDIF.
Assign the coefficients a_{2} through a_{n}. to vector a.
Calculate the interpolated value of Xint. The first term in the polynomial is a_{1}. The following terms are added by using a loop.
The NewtonsinT ( x, y, Xint) Function is then used in the Command Window for calculating the interpolated value of x=3.
>> x = [l 2 4 5 7];
>> y = [52 5 -5 -40 10) ;
>> Yinterpolated = NewtonsINT(x,y,3)
Yinterpolated =
6